Question

water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is 2m/s is 1.06×10^5pa.what will be the pressure in the portion where velocity is 4m/s
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Answer 1

Answer:

According to Bernoulli,s principle

 P1+21ρv12+ρgh1 =P2+21ρv22+ρgh2

⇒ 2mHg+21ρ(2)2 =P2+21ρ(4)2

⇒P2=21×103(22−42)+2mHg

⇒P2=−6000+2Hg

⇒−1013256000(0.76)mHg+2mHg

⇒P2=1.955 m of Hg